Answer
$\dfrac{e^6}{4}-\dfrac{7}{4}$
Work Step by Step
Let us consider that $u=x-y$ and $v=x+y$
$Jacobian, J(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=1 \cdot (1) -(-1)(1)=2$
and $J(u,v)=|\dfrac{1}{2}|$
Now, $\iint_R (x+y) e^{x^2-y^2} dA=\iint_{R} (x+y) e^{(x+y)(x-y)} dA= \int_0^{3} \int_0^2 (\dfrac{1}{2}) v e^{uv} du dv$
or, $=(\dfrac{1}{2}) \int_0^3 (e^{2v}-1) dv$
or, $=(\dfrac{1}{2}) [(\dfrac{1}{2}) (e^{2v}-v)]_0^3 $
or, $= \dfrac{e^6}{4}-\dfrac{7}{4}$
